科尔多瓦：拍照后上传图像不起作用

Question

我正在开发一个使用 Apache Cordova aka Phonegap 开发的 iOS 应用程序。 我想分两步上传照片： 1. 拍摄照片并以小尺寸显示照片 2.上传照片 我需要一键拍照和一键上传。

我的脚本不起作用。怎么了？

这是我的 JavaScript 文件：

var pictureSource;
var destinationType;

document.addEventListener("deviceready", onDeviceReady, false);

function onDeviceReady() {
    pictureSource = navigator.camera.PictureSourceType;
    destinationType = navigator.camera.DestinationType;
}

function clearCache() {
    navigator.camera.cleanup();
}

var retries = 0;
function onCapturePhoto(fileURI) {
    var win = function (r) {
        clearCache();
        retries = 0;
        navigator.notification.alert(
        '',
        onCapturePhoto,
        'Der Upload wurde abgeschlossen',
        'OK');
        console.log(r);
    }

    var fail = function (error) {
        navigator.notification.alert(
        'Bitte versuchen Sie es noch einmal.',
        onCapturePhoto,
        'Ein unerwarteter Fehler ist aufgetreten',
        'OK');
        console.log("upload error source " + error.source);
        console.log("upload error target " + error.target);
        if (retries == 0) {
            retries ++
            setTimeout(function() {
                onCapturePhoto(fileURI)
            }, 1000)
        } else {
            retries = 0;
            clearCache();
            alert('Fehler!');
        }
    }

    */do nothing*/
}


function capturePhoto() {
    navigator.camera.getPicture(onCapturePhoto, onFail, {
    quality: 50,
    destinationType: destinationType.FILE_URI
    });
}


function getPhoto(source) {
      navigator.camera.getPicture(onPhotoURISuccess, onFail, {
      quality: 50,
      destinationType: destinationType.FILE_URI,
      sourceType: source });
    }

function onFail(message) {
    alert('Failed because: ' + message);
}

function photoUpload(imageData) {
    var options = new FileUploadOptions();
    options.fileKey = "file";
    options.fileName = fileURI.substr(fileURI.lastIndexOf('/') + 1);
    options.mimeType = "image/jpeg";
    options.chunkedMode = false;

    var params = new Object();
    params.fileKey = "file";
    options.params = {}; // eig = params, if we need to send parameters to the server request


    var ft = new FileTransfer();
    ft.upload(fileURI, encodeURI("http://XXXXXXXX.com/app/upload.php"), win, fail, options);
}



<div id="camera">
    <button class="camera-control" onclick="capturePhoto();">Foto aufnehmen</button>
    <button class="camera-control" onclick="getPhoto(pictureSource.PHOTOLIBRARY);">From Photo Library</button><br>

    <div style="text-align:center;margin:20px;">
        <img id="cameraPic" src="" style="width:auto;height:120px;"></img>
    </div>

    <button class="camera-control" onclick="photoUpload(imageData);">UPLOAD</button>
</div>

Answer 1

更新：

我刚刚重构了您的代码，希望对您有所帮助。

JavaScript

<script> 
    var sPicData; //store image data for image upload functionality

    function capturePhoto(){
        navigator.camera.getPicture(picOnSuccess, picOnFailure, { 
            quality: 20,
            destinationType: Camera.DestinationType.DATA_URL,
            sourceType: Camera.PictureSourceType.CAMERA,
            correctOrientation: true
        });
    }

    function getPhoto(){
        navigator.camera.getPicture(picOnSuccess, picOnFailure, { 
            quality: 20,
            destinationType: Camera.DestinationType.DATA_URL,
            sourceType: Camera.PictureSourceType.SAVEDPHOTOALBUM,
            correctOrientation: true
        });
    }

    function picOnSuccess(imageData){

            var image = document.getElementById('cameraPic');
            image.src = imageData;
            sPicData  = imageData; //store image data in a variable
    }

    function picOnFailure(message){
        alert('Failed because: ' + message);
    }

    // ----- upload image ------------
    function photoUpload() {
        var options = new FileUploadOptions();
        options.fileKey = "file";
        options.fileName = sPicData.substr(sPicData.lastIndexOf('/') + 1);
        options.mimeType = "image/jpeg";
        options.chunkedMode = false;

        var params = new Object();
        params.fileKey = "file";
        options.params = {}; // eig = params, if we need to send parameters to the server request

        ft = new FileTransfer();
        ft.upload(sPicData, "http://XXXXXXXX.com/app/upload.php", win, fail, options); 
    }

    function win(){
        alert("image uploaded scuccesfuly");
    }

    function fail(){
        alert("image upload has been failed");
    }

</script>

HTML

<div id="camera">
    <button class="camera-control" onclick="capturePhoto();">Foto aufnehmen</button>
    <button class="camera-control" onclick="getPhoto();">From Photo Library</button><br>

    <div style="text-align:center;margin:20px;">
        <img id="cameraPic" src="" style="width:auto;height:120px;"></img>
    </div>

    <button class="camera-control" onclick="photoUpload();">UPLOAD</button>
</div>

Answer 2

1.拍摄照片并以小尺寸显示照片

• 在这里你可以设置成功的图片 onCapturePhoto(fileURI) e.g.

  <img id="cameraPic" src= "" style="width:120px;height:120px;" > </img> 
  
  function onCapturePhoto(fileURI) {
    $("#cameraPic").attr("src", fileURI);
  }
  

2。上传照片拍照一键上传

    <button class="camera-control" onclick="photoUpload();">UPLOAD</button>

    function photoUpload() {
        var fileURI = $("#cameraPic").attr("src");

        /* YOUR CODE TO UPLOAD IMAGE*/
    }